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C++中，static 成员函数可以表现出运行时多态性吗？

 3 years ago
source link: https://www.zhihu.com/question/440174887/answer/1687656318
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C++中，static 成员函数可以表现出运行时多态性吗？
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C++程序猿, 公众号：高级开发者
谢邀@IceBear

C++中内置的运行时多态特性即类的虚函数机制，若静态函数需要表现运行时多态性，得做额外的工作。这也是传说中的值语义多态，无需动态分配内存实现多态效果！

通过union来模拟加法类型Shape，使得静态函数通过统一的Shape类型，运行时根据不同的Subtype做出不同的算法：

struct Shape {
    enum { Circle, Retangle } type;
    union {
        struct {
            double r;
        } circle;
        struct {
            double w;
            double h;
        } retangle;
    };
};

struct Algorithm {
    static double area(const Shape& s) {
        switch (s.type) {
            case Shape::Circle:
                return 3.14 * s.circle.r * s.circle.r;
            case Shape::Retangle:
                return s.retangle.w * s.retangle.h;
        }
    }
};
Shape s {
    .type = Shape::Circle,
    .circle = {2}
};
std::cout << Algorithm::area(s) << std::endl;
s = {
    .type = Shape::Retangle,
    .retangle = {2, 3}
};
std::cout << Algorithm::area(s) << std::endl;

然而union是非类型安全的，并且无法直接与非平凡类组合，在实际中不那么实用了。可以采用C++17引入的std::variant来解决这个问题。

struct Circle {
    double r;
    double area() const {
        return 3.14* r * r;
    }
};

struct Retangle {
    double w;
    double h;
    double area() const {
        return w * h;
    }
};

using Shape = std::variant<Circle, Retangle>;
struct Algorithm {
    static double area(const Shape& s) {
        return std::visit([](const auto& c) {
            return c.area();
        }, s);
    };
};
Shape s = Circle{2};
std::cout << Algorithm::area(s) << std::endl;
s = Retangle{2, 3};
std::cout << Algorithm::area(s) << std::endl;

report

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